[Leetcode 237] Delete Node in a Linked List (easy)

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:

image

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

Solution

因为只给了需要删除的节点，并没有给头结点，所以是无法获取需要删除节点的前一个节点的，不能用prev.next = next这种方法。
为了删除节点，可以将需要删除节点后一个节点的值拷贝给删除节点，然后current.next = current.next.next 即可。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void deleteNode(ListNode node) {
        node.val = node.next.val;
        node.next = node.next.next;
    }
}